Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 11, Problem 11.41P

(a)

Interpretation Introduction

Interpretation:

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in IF2 is to be drawn.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.

(a)

Expert Solution
Check Mark

Answer to Problem 11.41P

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in IF2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  1

Explanation of Solution

The Lewis structure of IF2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  2

Iodine forms two single bonds with two fluorine atoms and three lone pairs are present on it so the hybridization of iodine in IF2 is sp3d.

The atomic number of iodine is 53 so its electronic configuration is [Kr]4d105s25p5.

The partial orbital diagram for an isolated I atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  3

The partial orbital for hybridized I atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  4

One s orbital, three p orbitals and one d orbital of central atom iodine combine to form five sp3d hybridized orbitals. Iodine has 7 valence electrons and it has one negative charge also that adds one more electron in the valence orbital of iodine in the orbital diagram. The formation of hybrid orbitals can be shown as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  5

Conclusion

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in IF2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  6

(b)

Interpretation Introduction

Interpretation:

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in ICl3 is to be drawn.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.

(b)

Expert Solution
Check Mark

Answer to Problem 11.41P

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in ICl3 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  7

Explanation of Solution

The Lewis structure of ICl3 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  8

Iodine forms three single bonds with three chlorine atoms and two lone pairs are present on it so five hybrid orbitals are required. The hybridization of I in ICl3 is sp3d.

The atomic number of iodine is 53 so its electronic configuration is [Kr]4d105s25p5.

The partial orbital diagram for an isolated I atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  9

The partial orbital for hybridized I atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  10

One s orbital, three p orbitals and one d orbital of central atom iodine combine to form five sp3d hybridized orbitals. The formation of hybrid orbitals can be shown as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  11

Conclusion

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom iodine in ICl3 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  12

(c)

Interpretation Introduction

Interpretation:

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in XeOF4 is to be drawn.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.

(c)

Expert Solution
Check Mark

Answer to Problem 11.41P

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom xenon in XeOF4 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  13

Explanation of Solution

The Lewis structure of XeOF4 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  14

Xenon forms four single bonds with four fluorine atoms and one double bond with oxygen and one lone pair is present on it so six hybrid orbitals are required. The hybridization of xenon in XeOF4 is sp3d2.

The atomic number of xenon is 54 so its electronic configuration is [Kr]4d105s25p6.

The partial orbital diagram for an isolated Xe atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  15

The partial orbital for hybridized Xe atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  16

One s orbital, three p orbitals and two d orbitals of central atom xenon combine to form six sp3d2 hybridized orbitals. The formation of hybrid orbitals can be shown as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  17

Conclusion

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom xenon in XeOF4 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  18

(d)

Interpretation Introduction

Interpretation:

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom in BHF2 is to be drawn.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

The partial orbital diagram is the one that shows the distribution of electrons in the valence shell only.

(d)

Expert Solution
Check Mark

Answer to Problem 11.41P

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom boron in BHF2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  19

Explanation of Solution

The Lewis structure of BHF2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  20

Boron forms one single bond with hydrogen and two single bonds with two fluorine atoms so three hybrid orbitals are required. The hybridization of boron in BHF2 is sp2.

The atomic number of boron is 5 so its electronic configuration is 1s22s22p1.

The partial orbital diagram for an isolated B atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  21

The partial orbital for hybridized B atom is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  22

One s orbital and two p orbitals of central atom boron combine to form three sp2 hybridized orbitals. The formation of hybrid orbitals can be shown as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  23

Conclusion

The partial orbital diagram that shows the formation of hybrid orbitals from the atomic orbitals of the central atom boron in BHF2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.41P , additional homework tip  24

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Chapter 11 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 11 - Prob. 11.3PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Give the number and type of hybrid orbital that...Ch. 11 - What is the hybridization of nitrogen in each of...Ch. 11 - What is the hybridization of carbon in each of the...Ch. 11 - Prob. 11.9PCh. 11 - Prob. 11.10PCh. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Phosphine (PH3) reacts with borane (BH3) as...Ch. 11 - The illustrations below depict differences in...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Methyl isocyanate, , is an intermediate in the...Ch. 11 - Are these statements true or false? Correct any...Ch. 11 - Prob. 11.21PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.23PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Certain atomic orbitals on two atoms were combined...Ch. 11 - Prob. 11.28PCh. 11 - Antibonding MOs always have at least one node. Can...Ch. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - The molecular orbitals depicted are derived from...Ch. 11 - The molecular orbitals depicted below are derived...Ch. 11 - Prob. 11.34PCh. 11 - Use an MO diagram and the bond order you obtain...Ch. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Epinephrine (or adrenaline; below) is a naturally...Ch. 11 - Prob. 11.41PCh. 11 - Isoniazid (below) is an antibacterial agent that...Ch. 11 - Prob. 11.43PCh. 11 - Prob. 11.44PCh. 11 - Prob. 11.45PCh. 11 - Prob. 11.46PCh. 11 - Tryptophan is one of the amino acids found in...Ch. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Prob. 11.51PCh. 11 - Prob. 11.52PCh. 11 - Sulfur forms oxides, oxoanions, and halides. What...Ch. 11 - Prob. 11.54PCh. 11 - Use an MO diagram to find the bond order and...Ch. 11 - Acetylsalicylic acid (aspirin), the most widely...Ch. 11 - Prob. 11.57P
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